Question

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Answer 1

Let r cm be the radius, h cm be the height, S cm² be the total surface area and V cm³ be the volume.

Now,

V = πr²h = 100

⇒ `h = 100/(pir^2)` 

and S = 2πr² + 2πrh

⇒ `S = 2pir^2 + 2pir (100/(pir^2))`

`= 2pir^2 + 200/r`

Differentiate `S = 2pir^2 + 200/r` w.r.t r we get

`(dS)/(dr) = 4pir - 200/r^2`

For maximum / minimum surface area 

`(dS)/(dr) = 0`

⇒ `4pir - 200/r^2 = 0`

⇒ `r^3 = 200/ (4pi)`

⇒ `r = (50/pi)^(1//3)`

`(d^2S)/(dr^2) = 4pi + (200 xx 2)/r^3`

`= 4pi + 400/r^3`

and `((d^2S)/(dr^2))_(r = (50/pi)^(1/3))`

`= 4pi + 400/ (50/pi) > 0`

∴ S has a minimum value at

`r = (50/pi)^(1/3)`

When `r = (50/pi)^(1/3)` cm, then

`h = 100/(pi(50/pi)^(2//3)) `

`h = 100/((50)^(2//3) pi^(1//3))`

`= (50xx2)/ ((50)^(2//3) pi ^(1//3))`

`= 2 (50/pi)^(1//3)` cm.

When r `(50/pi)^(1/3)` and h = 2 `(50/pi)^(1/3)` then S will be minimum.

Hence, the total surface area will be minimum.